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[SFW] Riddles / Puzzles / Brain Teasers - Page 17

Forum Index > General Forum
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Umpteen
Profile Blog Joined April 2010
United Kingdom1570 Posts
April 20 2012 09:22 GMT
#321
On April 20 2012 17:50 TheAngryZergling wrote:
Show nested quote +
On April 20 2012 07:06 Excludos wrote:
On April 20 2012 06:51 TheAngryZergling wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.


+ Show Spoiler +
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.


I address that point explicitly in my original solution in the following sentence:
+ Show Spoiler +

his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.


+ Show Spoiler +
No; sorry, you're missing the point.

You say that the counter adds 1 if the left lever is up, and flips it down.
You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers.
Everyone else flips the left lever up the first chance they get, but not subsequently.

Suppose the left lever starts down.
A non-counter enters the room first (unknowingly) and flicks the left lever up.
Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.

Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.

The existence of a food chain is inescapable if we evolved unsupervised, and inexcusable otherwise.
gameguard
Profile Blog Joined March 2006
Korea (South)2132 Posts
April 20 2012 09:28 GMT
#322
On April 20 2012 18:22 Umpteen wrote:
Show nested quote +
On April 20 2012 17:50 TheAngryZergling wrote:
On April 20 2012 07:06 Excludos wrote:
On April 20 2012 06:51 TheAngryZergling wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.


+ Show Spoiler +
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.


I address that point explicitly in my original solution in the following sentence:
+ Show Spoiler +

his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.


+ Show Spoiler +
No; sorry, you're missing the point.

You say that the counter adds 1 if the left lever is up, and flips it down.
You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers.
Everyone else flips the left lever up the first chance they get, but not subsequently.

Suppose the left lever starts down.
A non-counter enters the room first (unknowingly) and flicks the left lever up.
Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.

Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.



thats why someone said everyone flips it up twice
EngrishTeacher
Profile Blog Joined March 2012
Canada1109 Posts
April 20 2012 09:41 GMT
#323
On April 18 2012 23:44 Ghanburighan wrote:
[image loading]

+ Show Spoiler [Hint] +
Count something


+ Show Spoiler [solution] +
Count the number of circles in the numbers, for example 8 has two.


As a math/econ double major, I want to kill myself -_-

Spent almost 30 minutes doing all kinds of calculations, FML.
iTzSnypah
Profile Blog Joined February 2011
United States1738 Posts
April 20 2012 09:59 GMT
#324
Very easy riddle. ccccccc what am i?
Team Liquid needs more Terrans.
Umpteen
Profile Blog Joined April 2010
United Kingdom1570 Posts
April 20 2012 10:53 GMT
#325
On April 20 2012 18:28 gameguard wrote:
Show nested quote +
On April 20 2012 18:22 Umpteen wrote:
On April 20 2012 17:50 TheAngryZergling wrote:
On April 20 2012 07:06 Excludos wrote:
On April 20 2012 06:51 TheAngryZergling wrote:
On April 20 2012 02:33 XiGua wrote:
I am betting that nobody will solve this without cheating.

I present to you,
THE WARDEN

The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.

"In the prison is a switch room, which contains two light switches labeled 1 and 2, each of which can be in either up or the down position. I am not telling you their present positions. The switches are not connected to anything.

"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must flip one switch when he visits the switch room, and may only flip one of the switches. Then he'll be led back to his cell.

"No one else will be allowed to alter the switches until I lead the next prisoner into the switch room. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back. I will not touch the switches, if I wanted you dead you would already be dead.

"Given enough time, everyone will eventually visit the switch room the same number of times as everyone else. At any time, anyone may declare to me, 'We have all visited the switch room.'

"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will all die horribly. You will be carefully monitored, and any attempt to break any of these rules will result in instant death to all of you"

What is the strategy they come up with so that they can be free?


+ Show Spoiler +


there is a designated counter. if the left lever is up he adds 1 and flips it back down. if it is down he just toggles the right one. his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.
for everyone else you flip the left switch up on your first opportunity to do so. otherwise toggle the right switch.

when the counter has flipped the left switch down 22 more times after his initial visit. he tells the warden that they all have flipped the switches.


+ Show Spoiler +
Someone might come into the room first, not knowing if the counter has been there or not, and flips the switch up. If the counter then starts at 1 at his first visit, he will be stuck at 22 never reaching 23. The huge problem with this riddle is that you don't know the state of the switches before you enter the room.


I address that point explicitly in my original solution in the following sentence:
+ Show Spoiler +

his count is 1 when he leaves the room for the first time (himself) regardless of the state of the levers.


+ Show Spoiler +
No; sorry, you're missing the point.

You say that the counter adds 1 if the left lever is up, and flips it down.
You also say his count is 1 when he leaves the room for the first time, regardless of the state of the levers.
Everyone else flips the left lever up the first chance they get, but not subsequently.

Suppose the left lever starts down.
A non-counter enters the room first (unknowingly) and flicks the left lever up.
Some time later, the counter enters the room for the first time. The lever is up. You say he flips it down and leaves the room with a count of 1 (himself). But the non-counter who got there first will never flip the left lever again, so will never be counted. The counter can never reach 23.

Nor can the counter assume the above happened and start his count at 2 just because he found the lever up (since the lever might have started in the up position), because in that case he would declare victory one too soon.



+ Show Spoiler +
thats why someone said everyone flips it up twice


You need to spoiler that. And I know - but he wasn't saying that.
The existence of a food chain is inescapable if we evolved unsupervised, and inexcusable otherwise.
Forumite
Profile Joined February 2011
Sweden3280 Posts
April 20 2012 11:59 GMT
#326
On April 20 2012 16:43 Aelfric wrote:
Here is an easy one:

Show nested quote +
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?

+ Show Spoiler +
They tell their salary to a sixth person who they don´t work with, meaning he´s not a coworker.
:3
Prox
Profile Joined July 2010
Netherlands174 Posts
April 20 2012 12:18 GMT
#327
On April 20 2012 18:41 EngrishTeacher wrote:
Show nested quote +
On April 18 2012 23:44 Ghanburighan wrote:
[image loading]

+ Show Spoiler [Hint] +
Count something


+ Show Spoiler [solution] +
Count the number of circles in the numbers, for example 8 has two.


As a math/econ double major, I want to kill myself -_-

Spent almost 30 minutes doing all kinds of calculations, FML.


I solved it in 1-2 minutes. I didnt consider counting the circles but just the number combinations and total numbers.

0 = worth 1 because 0000 = 4
1 = worth 0 because 1111 = 0
2 = worth 0 because 2222 = 0
3 = worth 0 because 3333 = 0
4 is not in the problem
5 = worth 0 because 5555 = 0
6 = worth 1 because 6666 = 4
7 = worth 0 because 7777 = 0
9 = worth 1 because 9999 = 4

8 = worth 2 because 8809 = 2 + 2 + 1 + 1 = 6
and to double check it corresponds with 8193 = 2 + 0 + 1 + 0 = 3
or 9881 = 1 + 2 + 2 + 0 = 5

so the final one is 2581 = 0 + 0 + 2 + 0 = 2

Obviously I didnt notice the corelation between circles and numbers. Funny problem.

It would be alot more fun and harder if they left out a couple of combinations, like 5555 2222 and 1111.
http://www.teamliquid.net/video/streams/Prox
ThePlayer33
Profile Joined October 2011
Australia2378 Posts
April 20 2012 12:39 GMT
#328
On April 20 2012 21:18 Prox wrote:
Show nested quote +
On April 20 2012 18:41 EngrishTeacher wrote:
On April 18 2012 23:44 Ghanburighan wrote:
[image loading]

+ Show Spoiler [Hint] +
Count something


+ Show Spoiler [solution] +
Count the number of circles in the numbers, for example 8 has two.


As a math/econ double major, I want to kill myself -_-

Spent almost 30 minutes doing all kinds of calculations, FML.


I solved it in 1-2 minutes. I didnt consider counting the circles but just the number combinations and total numbers.

0 = worth 1 because 0000 = 4
1 = worth 0 because 1111 = 0
2 = worth 0 because 2222 = 0
3 = worth 0 because 3333 = 0
4 is not in the problem
5 = worth 0 because 5555 = 0
6 = worth 1 because 6666 = 4
7 = worth 0 because 7777 = 0
9 = worth 1 because 9999 = 4

8 = worth 2 because 8809 = 2 + 2 + 1 + 1 = 6
and to double check it corresponds with 8193 = 2 + 0 + 1 + 0 = 3
or 9881 = 1 + 2 + 2 + 0 = 5

so the final one is 2581 = 0 + 0 + 2 + 0 = 2

Obviously I didnt notice the corelation between circles and numbers. Funny problem.

It would be alot more fun and harder if they left out a couple of combinations, like 5555 2222 and 1111.


agreed! would been fun to just slide 1 or 2 of the quadruple digits in and leave the rest out.
| Idra | YuGiOh | Leenock | Coca |
dmfg
Profile Joined May 2008
United Kingdom591 Posts
April 20 2012 12:46 GMT
#329
On April 20 2012 18:14 iTzSnypah wrote:
Show nested quote +
On April 20 2012 18:13 TheAngryZergling wrote:
On April 20 2012 17:59 iTzSnypah wrote:
ABOUT the Warden Puzzle: The OP did not include 1 Crucial statement.
+ Show Spoiler [THIS IS THE ANSWER] +
http://www.braingle.com/brainteasers/teaser.php?op=2&id=18012&comm=0


please provide the crucial statement, not only the answer which I would rather not see.


Switches are down position to begin with.


I think part of the point of this riddle is that you don't know what the initial position of the switches are.

There's another version of this riddle where you do know the starting position, but there is only a single switch.

FWIW I can't remember the solution to either ^^ but I think they're equivalent
dmfg
Profile Joined May 2008
United Kingdom591 Posts
Last Edited: 2012-04-20 13:20:57
April 20 2012 13:06 GMT
#330
On April 20 2012 16:43 Aelfric wrote:
Here is an easy one:

Show nested quote +
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?


OK, let's have a go at this. Gonna try a 3 person version first and see where I get.

+ Show Spoiler +
Let's call these people a, b and c with salaries A, B and C.

Then a needs to find out (B + C) without ever being able to individually deduce B or C, b needs to find out (A + C), etc.

Then b pretty much has to give some altered (but undecipherable) form of his salary to c, so c can pass the (altered) sum on to a. The complete sum has to come back to b to be deciphered, since otherwise one of the others would be able to use the information to deduce his salary.

So, let's have a, b and c generate some arbitrary numbers and hand them out.
- a thinks of a number D and tells b only
- b thinks of a number E and tells c only
- c thinks of a number F and tells a only

Then each person multiplies their salary by the number and gives it to the third person:
- a tells AF to b (who doesn't know F)
- b tells BD to c (who doesn't know D)
- c tells CE to a (who doesn't know E)

Now we can give salted sums of 2 salaries for the 3rd person to add his. Although the 3rd person knows the multiplier, he's given a sum of 2 salaries so can cannot work out either.
- a tells (A + CE) to b
- b tells (B + AF) to c
- c tells (C + BD) to a

We can now get people to put together all 3 salaries, and tell this sum to the person with the modified salary in the sum to decipher:
- a tells (A + BD + C) to b
- b tells (A + B + CE) to c
- c tells (AF + B + C) to a

Now each of the people can subtract (their own salary * the multiplier they were given at the start), add their own salary, and divide by 3 to get the average.

At no point is anyone able to calculate either of the other two people's salaries.

Does this same thing work for 5 people?


EDIT: Let's try 5 people then.

+ Show Spoiler +
Let's call these people a, b, c, d and e with salaries A, B, C, D and E.

So, let's have a, b and c generate some arbitrary numbers and hand them out.
- a thinks of a number F and tells b only
- b thinks of a number G and tells c only
- c thinks of a number H and tells d only
- d thinks of a number I and tells e only
- e thinks of a number J and tells a only

Then each person multiplies their salary by the number and gives it to the next person:
- a tells AJ to b (who doesn't know J)
- b tells BF to c (who doesn't know F)
- c tells CG to d (who doesn't know G)
- d tells DH to e (who doesn't know H)
- e tells EI to a (who doesn't know I)

Now we can give salted sums of 2 salaries for a 3rd person to add his.
- a tells (A + EI) to b
- b tells (B + AJ) to c
- c tells (C + BF) to d
- d tells (D + CG) to e
- e tells (E + DH) to a

And again each person adds their salary and passes it around again, for a 4th person to add his.
- a tells (A + DH + E) to b
- b tells (A + B + EI) to c
- c tells (AJ + B + C) to d
- d tells (BF + C + D) to e
- e tells (CG + D + E) to a

And again.
- a tells (A + CG + D + E) to b
- b tells (A + B + DH + E) to c
- c tells (A + B + C + EI) to d
- d tells (AJ + B + C + D) to e
- e tells (BF + C + D + E) to a

One final time, to pass the final sum around. The final sum arrives at the person whose own salary is altered, so they can decipher it.
- a tells (A + BF + C + D + E) to b
- b tells (A + B + CG + D + E) to c
- c tells (A + B + C + DH + E) to d
- d tells (A + B + C + D + EI) to e
- e tells (AJ + B + C + D + E) to a

Now each of the people can subtract (their own salary * the multiplier they were given at the start), add their own salary, and divide by 5 to get the average.

At no point is anyone able to calculate any other the other people's salaries.


EDIT2: That's actually a really convoluted way of doing it wasn't it!! Simple solution below

+ Show Spoiler +
Thinking about it, was it completely unnecessary to have someone else generate a multiplier in the first step?

All we actually need is for each person to generate a random number, get everyone else to add their salary to it in turn, and then when it gets back to each person only they can subtract their random number and add their salary to get the total :p
awu25
Profile Joined April 2010
United States2003 Posts
April 20 2012 13:25 GMT
#331
On April 20 2012 18:41 EngrishTeacher wrote:
Show nested quote +
On April 18 2012 23:44 Ghanburighan wrote:
[image loading]

+ Show Spoiler [Hint] +
Count something


+ Show Spoiler [solution] +
Count the number of circles in the numbers, for example 8 has two.


As a math/econ double major, I want to kill myself -_-

Spent almost 30 minutes doing all kinds of calculations, FML.

I hate how whoever created this puzzle included the preschool thing. I highly doubt it's true
Umpteen
Profile Blog Joined April 2010
United Kingdom1570 Posts
April 20 2012 13:26 GMT
#332
On April 20 2012 21:46 dmfg wrote:
Show nested quote +
On April 20 2012 18:14 iTzSnypah wrote:
On April 20 2012 18:13 TheAngryZergling wrote:
On April 20 2012 17:59 iTzSnypah wrote:
ABOUT the Warden Puzzle: The OP did not include 1 Crucial statement.
+ Show Spoiler [THIS IS THE ANSWER] +
http://www.braingle.com/brainteasers/teaser.php?op=2&id=18012&comm=0


please provide the crucial statement, not only the answer which I would rather not see.


Switches are down position to begin with.


I think part of the point of this riddle is that you don't know what the initial position of the switches are.

There's another version of this riddle where you do know the starting position, but there is only a single switch.

FWIW I can't remember the solution to either ^^ but I think they're equivalent


Yeah, it is do-able without knowing the initial switch positions, so the info isn't missing.
The existence of a food chain is inescapable if we evolved unsupervised, and inexcusable otherwise.
Aragnis
Profile Joined April 2011
Australia17 Posts
April 20 2012 13:30 GMT
#333
On April 20 2012 16:43 Aelfric wrote:
Here is an easy one:

Show nested quote +
Five coworkers want to know what the average of all their salaries is, but refuse to reveal ANY information about their own salaries to their coworkers. How can they calculate the average?


(Working out the average of salaries without telling anyone what your salary is)
+ Show Spoiler +

The simplest way is to find a trusted 3rd party who promises not to tell anyone what anyone else's salary is, have everyone tell that person their salary, and they work out the average and let everyone know what it is. But if you can't find such a person, there's another way...

Arrange the cow-orkers in a line, and have them all pick a random number and remember it (without telling anyone else). This number is allowed to be negative, and can be as big as they like.

The first person adds his salary and random number together, and tells that sum to the next person (without telling anyone else). They add on their salary and their random number, and tell the new total to the next person, who does the same, through all of them.

The last person, after adding on her salary and random number, tells her result to the first person.

The first person then subtracts his random number from the number he was told, and passes the answer on to the second person. They subtract their random number, pass the result on, down the line again. The last person, after subtracting her random number, announces the number to everybody. That is the total salary of them all, and they just have to divide by the number of people to get the average.

For the first person, the only information they get passed is the sum of everyone's salary, obscured by all their random numbers. He can't figure anything out from that.

For everyone else except the last person, the two numbers they are told are obscured either by the random numbers of the people in front of them (the first time through) or after them (the second time through). So they can't tell anything either.

The last person has nobody after them, so the number they get the second time through tells them something! Well no, not really, since all it tells them is the total salary of everyone else, and everyone else can work out the same thing once they are told the total salary, just by subtracting their own salary. So the system does not reveal any additional information about anyone else's salary to anyone. QED

(however, the system is not perfect. If the people standing on either side of someone get together, they can work out that person's salary without revealing to each other anything more about their own salaries.)

Sandster
Profile Joined November 2006
United States4054 Posts
Last Edited: 2012-04-20 16:14:23
April 20 2012 15:40 GMT
#334
Horse question solution. Haven't looked this up or anything but pretty confident.
+ Show Spoiler +
9 races. I'll break them into lettered groups.

A: 5 races of 5 horses each. Group stages.
B: 1 race of 5 winners of group stages A. Winner is your fastest horse.
C: 1 race of runner ups of A.
D: 1 race of 3rd place finishers of A.
E: 1 race to determine 2nd/3rd fastest horses:
-2nd and 3rd place of race B
-1st and 2nd place of race C
-1st place of race D

Rationale:
-Race B winner is fastest horse, so you only have to worry about the other 2 slots.
-Race D: to qualify, the horse finished 3rd in its original group, meaning 2 horses were already faster. Therefore only the winner in Race D has a chance to be among the top 3 overall (or it'd be at best 4th, behind the 1st/2nd place in its race A, and 1st in race D).
-Similar rationale for Race C: only the 1st/2nd place finishers have a chance to be top 3 overall, or else they'd be at best 4th (behind 1st place in their race A, and 1st/2nd in race C).
-You have exactly 5 potential horses remaining for race E to determine your 2nd/3rd fastest horses overall.

EDIT: This assumes you couldn't identify horses across races. If you could, it would be easier, and take 7 races.

Same A and B as above. Then one race of:
-2nd and 3rd place of race B
-2nd and 3rd place of the race A that had the horse who won race B
-2nd place of the race A that had the horse who was 2nd in race B

Reason here being if you finished 2nd/3rd in a race where the winner wasn't the top 3, there's no chance that you're one of the top 3.

Aelfric
Profile Blog Joined March 2010
Turkey1496 Posts
Last Edited: 2012-04-20 16:39:39
April 20 2012 16:36 GMT
#335
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution.
It may have been already asked, but here is a new one to keep you busy:


[image loading]

Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.

Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
Tomorrow never comes until its too late...
ZasZ.
Profile Joined May 2010
United States2911 Posts
April 20 2012 16:53 GMT
#336
On April 21 2012 01:36 Aelfric wrote:
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution.
It may have been already asked, but here is a new one to keep you busy:

Show nested quote +

[image loading]

Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.

Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?


+ Show Spoiler +
You don't spin the barrel. If you spin it, you have a 1/3 chance of landing on a chamber with a bullet, and 2/3 chance of landing on a chamber without one. If you don't spin it, you have a 1/4 chance of the next chamber having a bullet, and a 3/4 chance of the next chamber being empty. 3/4 > 2/3.
MaestroSC
Profile Blog Joined August 2009
United States2073 Posts
April 20 2012 17:03 GMT
#337
On April 21 2012 01:36 Aelfric wrote:
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution.
It may have been already asked, but here is a new one to keep you busy:

Show nested quote +

[image loading]

Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.

Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?


+ Show Spoiler +
Statistically, you should not spin it. The chance is 1/4 that the first shot was in the position near the 2 consecutive bullets. There is a 3/4 or 75% chance that it was in locations 1-3 of the 4 consecutive empty slots. Some people are going to argue for the slots that are filled with bullets should be taken into the equation, but knowing that the first shot was a miss means that you know you are dealing with bullet slots 1-5 as the very worst spot the bullet could be at this moment is in slot number 5. there is only a 25% chance it is there. Because you know as soon as the first shot doesnt go off that it its NOT in slot 1. Therefore it can only be slots 2-5.

If you spin the gun you chance of survival drops from 75% to 66%. Its a 6 round chamber, and 2/6 or 1/3 are bullets. You are giving an opportunity for it to land on any of the 6. Whereas before there is only 1 bullet to 3 empty slots possible.
NTTemplar
Profile Joined August 2011
609 Posts
Last Edited: 2012-04-20 17:08:40
April 20 2012 17:03 GMT
#338
On April 21 2012 01:36 Aelfric wrote:
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution.
It may have been already asked, but here is a new one to keep you busy:

Show nested quote +

[image loading]

Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.

Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?


+ Show Spoiler +
Edited, was wrong T_T person above was right
"Between Tomorrow's dream and yesterday's regret, is today's opportunity"
ZasZ.
Profile Joined May 2010
United States2911 Posts
Last Edited: 2012-04-20 17:09:40
April 20 2012 17:07 GMT
#339
On April 21 2012 02:03 NTTemplar wrote:
Show nested quote +
On April 21 2012 01:36 Aelfric wrote:
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution.
It may have been already asked, but here is a new one to keep you busy:


[image loading]

Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.

Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?


+ Show Spoiler +
You should spin it, at the moment you got 3/5 safe and 2/5 unsafe, if you spin it you get 4/6 safe 2/6 unsafe.

Having pulled and alleready used up one safe it is better to restart it than continue with one less guaranteed safe slot.


+ Show Spoiler +
Your rationale only applies if the chamber the gun fires from is random, but it isn't. It always fires the next bullet in the chamber.
NTTemplar
Profile Joined August 2011
609 Posts
April 20 2012 17:09 GMT
#340
On April 21 2012 02:07 ZasZ. wrote:
Show nested quote +
On April 21 2012 02:03 NTTemplar wrote:
On April 21 2012 01:36 Aelfric wrote:
I'm sending the detailed answer of Horse Race question via pm to whoever gets close to the solution.
It may have been already asked, but here is a new one to keep you busy:


[image loading]

Your enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.

Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?


+ Show Spoiler +
You should spin it, at the moment you got 3/5 safe and 2/5 unsafe, if you spin it you get 4/6 safe 2/6 unsafe.

Having pulled and alleready used up one safe it is better to restart it than continue with one less guaranteed safe slot.


Your rationale only applies if the chamber the gun fires from is random, but it isn't. It always fires the next bullet in the chamber.


Yeah noticed, edited it.

btw you should spoiler that
"Between Tomorrow's dream and yesterday's regret, is today's opportunity"
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